Maximum angle between vectors
Consider 3 vectors $textbf{v},textbf{v}',textbf{u}$ related by
$$textbf{v}=textbf{u}+textbf{v}'$$
Let $theta$ be the angle between $textbf{v}$ and $textbf{u}$ and let $phi$ be the angle between $textbf{v}$ and $-textbf{v}'$.
For what angle $theta$ is the angle $phi$ a maximum? The magnitudes $v,u$ are given and you can express the angle in terms of them.
I started by doing the dot product of the equation with itself getting
$$v^{2}=vucostheta+vv'cosphi=u^{2}+v^{'2}+2uv'cos (pi/2-phi-theta)$$
I thought I would then find the derivative of this expression with respect to $theta$ and set $dphi/dtheta=0$. That gives
$$-vusintheta=2uv'sin(pi/2-phi-theta)$$ or
$$vsintheta=2v'sin(phi+theta-pi/2)=-2v'cos(phi+theta)$$
which seems overdetermined.
calculus vectors
asked 5 hours ago
user2175783
1575
add a comment |
Consider 3 vectors $textbf{v},textbf{v}',textbf{u}$ related by
$$textbf{v}=textbf{u}+textbf{v}'$$
Let $theta$ be the angle between $textbf{v}$ and $textbf{u}$ and let $phi$ be the angle between $textbf{v}$ and $-textbf{v}'$.
For what angle $theta$ is the angle $phi$ a maximum? The magnitudes $v,u$ are given and you can express the angle in terms of them.
I started by doing the dot product of the equation with itself getting
$$v^{2}=vucostheta+vv'cosphi=u^{2}+v^{'2}+2uv'cos (pi/2-phi-theta)$$
I thought I would then find the derivative of this expression with respect to $theta$ and set $dphi/dtheta=0$. That gives
$$-vusintheta=2uv'sin(pi/2-phi-theta)$$ or
$$vsintheta=2v'sin(phi+theta-pi/2)=-2v'cos(phi+theta)$$
which seems overdetermined.
calculus vectors
asked 5 hours ago
user2175783
1575
1
Lenghts are given?
– greedoid
5 hours ago
On geometric grounds, if $|u|<|v'|$, then $theta$ should be $0$ or $pi$ such that $v$ and $v'$ are opposite ...
– Henning Makholm
5 hours ago
2
If all three magnitudes are given, then both $theta$ and $phi$ can be computed by standard triangle solving. There's no choice and no maxima or minima.
– Henning Makholm
5 hours ago
@HenningMakholm Only v,u are given (not v'). I edited the question.
– user2175783
5 hours ago
add a comment |
Consider 3 vectors $textbf{v},textbf{v}',textbf{u}$ related by
$$textbf{v}=textbf{u}+textbf{v}'$$
Let $theta$ be the angle between $textbf{v}$ and $textbf{u}$ and let $phi$ be the angle between $textbf{v}$ and $-textbf{v}'$.
For what angle $theta$ is the angle $phi$ a maximum? The magnitudes $v,u$ are given and you can express the angle in terms of them.
I started by doing the dot product of the equation with itself getting
$$v^{2}=vucostheta+vv'cosphi=u^{2}+v^{'2}+2uv'cos (pi/2-phi-theta)$$
I thought I would then find the derivative of this expression with respect to $theta$ and set $dphi/dtheta=0$. That gives
$$-vusintheta=2uv'sin(pi/2-phi-theta)$$ or
$$vsintheta=2v'sin(phi+theta-pi/2)=-2v'cos(phi+theta)$$
which seems overdetermined.
calculus vectors
asked 5 hours ago
user2175783
1575
Consider 3 vectors $textbf{v},textbf{v}',textbf{u}$ related by
$$textbf{v}=textbf{u}+textbf{v}'$$
Let $theta$ be the angle between $textbf{v}$ and $textbf{u}$ and let $phi$ be the angle between $textbf{v}$ and $-textbf{v}'$.
For what angle $theta$ is the angle $phi$ a maximum? The magnitudes $v,u$ are given and you can express the angle in terms of them.
I started by doing the dot product of the equation with itself getting
$$v^{2}=vucostheta+vv'cosphi=u^{2}+v^{'2}+2uv'cos (pi/2-phi-theta)$$
I thought I would then find the derivative of this expression with respect to $theta$ and set $dphi/dtheta=0$. That gives
$$-vusintheta=2uv'sin(pi/2-phi-theta)$$ or
$$vsintheta=2v'sin(phi+theta-pi/2)=-2v'cos(phi+theta)$$
which seems overdetermined.
calculus vectors
calculus vectors
asked 5 hours ago
user2175783
1575
asked 5 hours ago
user2175783
1575
edited 5 hours ago
asked 5 hours ago
user2175783
1575
asked 5 hours ago
user2175783
1575
asked 5 hours ago
user2175783
1575
1575
1
Lenghts are given?
– greedoid
5 hours ago
On geometric grounds, if $|u|<|v'|$, then $theta$ should be $0$ or $pi$ such that $v$ and $v'$ are opposite ...
– Henning Makholm
5 hours ago
2
If all three magnitudes are given, then both $theta$ and $phi$ can be computed by standard triangle solving. There's no choice and no maxima or minima.
– Henning Makholm
5 hours ago
@HenningMakholm Only v,u are given (not v'). I edited the question.
– user2175783
5 hours ago
add a comment |
1
Lenghts are given?
– greedoid
5 hours ago
On geometric grounds, if $|u|<|v'|$, then $theta$ should be $0$ or $pi$ such that $v$ and $v'$ are opposite ...
– Henning Makholm
5 hours ago
2
If all three magnitudes are given, then both $theta$ and $phi$ can be computed by standard triangle solving. There's no choice and no maxima or minima.
– Henning Makholm
5 hours ago
@HenningMakholm Only v,u are given (not v'). I edited the question.
– user2175783
5 hours ago
1
1
Lenghts are given?
– greedoid
5 hours ago
Lenghts are given?
– greedoid
5 hours ago
On geometric grounds, if $|u|<|v'|$, then $theta$ should be $0$ or $pi$ such that $v$ and $v'$ are opposite ...
– Henning Makholm
5 hours ago
On geometric grounds, if $|u|<|v'|$, then $theta$ should be $0$ or $pi$ such that $v$ and $v'$ are opposite ...
– Henning Makholm
5 hours ago
2
2
If all three magnitudes are given, then both $theta$ and $phi$ can be computed by standard triangle solving. There's no choice and no maxima or minima.
– Henning Makholm
5 hours ago
If all three magnitudes are given, then both $theta$ and $phi$ can be computed by standard triangle solving. There's no choice and no maxima or minima.
– Henning Makholm
5 hours ago
@HenningMakholm Only v,u are given (not v'). I edited the question.
– user2175783
5 hours ago
@HenningMakholm Only v,u are given (not v'). I edited the question.
– user2175783
5 hours ago
add a comment |
2 Answers
2
active
oldest
votes
Unless you're working under peculiar constraints, doing this algebraically is almost certainly a tactical error. Instead, draw a diagram:
A
/
u/ v
|/ |
B------>C
v'
The angles are $phi$ at C and $theta$ at A.
$phi$ will certainly be a maximum if we can make it $pi$ which happens if $|u|>|v|$ and $theta=0$.
Then consider the case $|u|<|v|$. If we have already decided on the positions of A and C, the possible positions of B is a circle with center A. Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle -- which are the ones where the angle is B is right!
Basic trigonometry then gives us $|u|=|v|costheta$.
answered 5 hours ago
Henning Makholm
237k16301536
"Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
– user35202
3 hours ago
1
@user35202: Geometrically, the tangent to the circle through C separates the rays that intersect the circle (which are the ones that represent possible $phi$s) from those that don't intersect the circle (and therefor cannot be possible $phi$s).
– Henning Makholm
3 hours ago
add a comment |
$$
vec v = vec u -(-vec u') = vec v = vec u -vec w
$$
so now
$$
vec ucdotvec v = ||vec u||^2-vec ucdotvec w
$$
or
$$
||vec u||||vec v||costheta=||vec u|| - ||vec u||||vec w||cosphi
$$
and then assuming that $vec ucdotvec w ne 0$
$$
phi = arccosleft(a+bcosthetaright)
$$
and now deriving
$$
frac{dphi}{dtheta} = frac{b sin (theta )}{sqrt{1-(a+b cos (theta ))^2}} = 0
$$
which gives $theta = 0 + kpi$
answered 5 hours ago
Cesareo
8,0883516
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3052520%2fmaximum-angle-between-vectors%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Unless you're working under peculiar constraints, doing this algebraically is almost certainly a tactical error. Instead, draw a diagram:
A
/
u/ v
|/ |
B------>C
v'
The angles are $phi$ at C and $theta$ at A.
$phi$ will certainly be a maximum if we can make it $pi$ which happens if $|u|>|v|$ and $theta=0$.
Then consider the case $|u|<|v|$. If we have already decided on the positions of A and C, the possible positions of B is a circle with center A. Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle -- which are the ones where the angle is B is right!
Basic trigonometry then gives us $|u|=|v|costheta$.
answered 5 hours ago
Henning Makholm
237k16301536
"Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
– user35202
3 hours ago
1
@user35202: Geometrically, the tangent to the circle through C separates the rays that intersect the circle (which are the ones that represent possible $phi$s) from those that don't intersect the circle (and therefor cannot be possible $phi$s).
– Henning Makholm
3 hours ago
add a comment |
Unless you're working under peculiar constraints, doing this algebraically is almost certainly a tactical error. Instead, draw a diagram:
A
/
u/ v
|/ |
B------>C
v'
The angles are $phi$ at C and $theta$ at A.
$phi$ will certainly be a maximum if we can make it $pi$ which happens if $|u|>|v|$ and $theta=0$.
Then consider the case $|u|<|v|$. If we have already decided on the positions of A and C, the possible positions of B is a circle with center A. Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle -- which are the ones where the angle is B is right!
Basic trigonometry then gives us $|u|=|v|costheta$.
answered 5 hours ago
Henning Makholm
237k16301536
"Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
– user35202
3 hours ago
1
@user35202: Geometrically, the tangent to the circle through C separates the rays that intersect the circle (which are the ones that represent possible $phi$s) from those that don't intersect the circle (and therefor cannot be possible $phi$s).
– Henning Makholm
3 hours ago
add a comment |
Unless you're working under peculiar constraints, doing this algebraically is almost certainly a tactical error. Instead, draw a diagram:
A
/
u/ v
|/ |
B------>C
v'
The angles are $phi$ at C and $theta$ at A.
$phi$ will certainly be a maximum if we can make it $pi$ which happens if $|u|>|v|$ and $theta=0$.
Then consider the case $|u|<|v|$. If we have already decided on the positions of A and C, the possible positions of B is a circle with center A. Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle -- which are the ones where the angle is B is right!
Basic trigonometry then gives us $|u|=|v|costheta$.
answered 5 hours ago
Henning Makholm
237k16301536
Unless you're working under peculiar constraints, doing this algebraically is almost certainly a tactical error. Instead, draw a diagram:
A
/
u/ v
|/ |
B------>C
v'
The angles are $phi$ at C and $theta$ at A.
$phi$ will certainly be a maximum if we can make it $pi$ which happens if $|u|>|v|$ and $theta=0$.
Then consider the case $|u|<|v|$. If we have already decided on the positions of A and C, the possible positions of B is a circle with center A. Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle -- which are the ones where the angle is B is right!
Basic trigonometry then gives us $|u|=|v|costheta$.
answered 5 hours ago
Henning Makholm
237k16301536
answered 5 hours ago
Henning Makholm
237k16301536
answered 5 hours ago
Henning Makholm
237k16301536
answered 5 hours ago
Henning Makholm
237k16301536
237k16301536
"Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
– user35202
3 hours ago
1
@user35202: Geometrically, the tangent to the circle through C separates the rays that intersect the circle (which are the ones that represent possible $phi$s) from those that don't intersect the circle (and therefor cannot be possible $phi$s).
– Henning Makholm
3 hours ago
add a comment |
"Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
– user35202
3 hours ago
1
@user35202: Geometrically, the tangent to the circle through C separates the rays that intersect the circle (which are the ones that represent possible $phi$s) from those that don't intersect the circle (and therefor cannot be possible $phi$s).
– Henning Makholm
3 hours ago
"Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
– user35202
3 hours ago
"Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
– user35202
3 hours ago
1
1
@user35202: Geometrically, the tangent to the circle through C separates the rays that intersect the circle (which are the ones that represent possible $phi$s) from those that don't intersect the circle (and therefor cannot be possible $phi$s).
– Henning Makholm
3 hours ago
@user35202: Geometrically, the tangent to the circle through C separates the rays that intersect the circle (which are the ones that represent possible $phi$s) from those that don't intersect the circle (and therefor cannot be possible $phi$s).
– Henning Makholm
3 hours ago
add a comment |
$$
vec v = vec u -(-vec u') = vec v = vec u -vec w
$$
so now
$$
vec ucdotvec v = ||vec u||^2-vec ucdotvec w
$$
or
$$
||vec u||||vec v||costheta=||vec u|| - ||vec u||||vec w||cosphi
$$
and then assuming that $vec ucdotvec w ne 0$
$$
phi = arccosleft(a+bcosthetaright)
$$
and now deriving
$$
frac{dphi}{dtheta} = frac{b sin (theta )}{sqrt{1-(a+b cos (theta ))^2}} = 0
$$
which gives $theta = 0 + kpi$
answered 5 hours ago
Cesareo
8,0883516
add a comment |
$$
vec v = vec u -(-vec u') = vec v = vec u -vec w
$$
so now
$$
vec ucdotvec v = ||vec u||^2-vec ucdotvec w
$$
or
$$
||vec u||||vec v||costheta=||vec u|| - ||vec u||||vec w||cosphi
$$
and then assuming that $vec ucdotvec w ne 0$
$$
phi = arccosleft(a+bcosthetaright)
$$
and now deriving
$$
frac{dphi}{dtheta} = frac{b sin (theta )}{sqrt{1-(a+b cos (theta ))^2}} = 0
$$
which gives $theta = 0 + kpi$
answered 5 hours ago
Cesareo
8,0883516
add a comment |
$$
vec v = vec u -(-vec u') = vec v = vec u -vec w
$$
so now
$$
vec ucdotvec v = ||vec u||^2-vec ucdotvec w
$$
or
$$
||vec u||||vec v||costheta=||vec u|| - ||vec u||||vec w||cosphi
$$
and then assuming that $vec ucdotvec w ne 0$
$$
phi = arccosleft(a+bcosthetaright)
$$
and now deriving
$$
frac{dphi}{dtheta} = frac{b sin (theta )}{sqrt{1-(a+b cos (theta ))^2}} = 0
$$
which gives $theta = 0 + kpi$
answered 5 hours ago
Cesareo
8,0883516
$$
vec v = vec u -(-vec u') = vec v = vec u -vec w
$$
so now
$$
vec ucdotvec v = ||vec u||^2-vec ucdotvec w
$$
or
$$
||vec u||||vec v||costheta=||vec u|| - ||vec u||||vec w||cosphi
$$
and then assuming that $vec ucdotvec w ne 0$
$$
phi = arccosleft(a+bcosthetaright)
$$
and now deriving
$$
frac{dphi}{dtheta} = frac{b sin (theta )}{sqrt{1-(a+b cos (theta ))^2}} = 0
$$
which gives $theta = 0 + kpi$
answered 5 hours ago
Cesareo
8,0883516
answered 5 hours ago
Cesareo
8,0883516
answered 5 hours ago
Cesareo
8,0883516
answered 5 hours ago
Cesareo
8,0883516
8,0883516
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3052520%2fmaximum-angle-between-vectors%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
Lenghts are given?
– greedoid
5 hours ago
On geometric grounds, if $|u|<|v'|$, then $theta$ should be $0$ or $pi$ such that $v$ and $v'$ are opposite ...
– Henning Makholm
5 hours ago
2
If all three magnitudes are given, then both $theta$ and $phi$ can be computed by standard triangle solving. There's no choice and no maxima or minima.
– Henning Makholm
5 hours ago
@HenningMakholm Only v,u are given (not v'). I edited the question.
– user2175783
5 hours ago